Suppose that an object is thrown down towards the ground from the top of the 1100 foot tall sears tower in Chicago with an initial velocity of -20 feet per second.Given that the acceleration of the object is constant at
-32 feet per second,determine how far above the ground the object is exactly six seconds after being thrown?

Nov 24th 2009, 04:06 PM

qmech

Kinematics

Remember the distance formula:

s = 0.5at^2 + vt + s0?

Nov 24th 2009, 04:15 PM

mathcalculushelp

Distance formula

Could you pls elaborate on that?

All I could understand from the problem is that

v(t)=-20ft/sec
a(t)=-32ft/sec
h=1100 foot
t=6secs

How does distance formula used in this?Iam confused(Doh)

Nov 24th 2009, 04:21 PM

qmech

Your textbook...

Your text should have a section entitled:

Kinematics
Constant Velocity Motion
Uniformly accelerated Motion
or some such title. See the examples in this section.

Your problem on free fall is motion under constant acceleration with a non-zero initial velocity.

Nov 24th 2009, 06:32 PM

Jameson

The kinematic formulas can be derived with calculus, but are easy to remember.

Start with a(t)=-32. Integrate to find v(t) and use the info that v(0)=-20 to solve for the +C that results from the integral. Integrate again to get s(t) and use s(0)=1100 to solve for the +C again.

OP - You need to show effort in your posts. If this all seems new to you then your teacher hasn't covered enough material for the problem. Does this make sense to you?